conduction heat transfer laboratory experiment - Aeronautical Engineering.

1. Student Name: Ali Uday Harobey
Lecturer Name: Dr. Basim Abdulrazzak
Conduction
Heat transfer laboratory
Aeronautical Engineering Department
-University of Baghdad

2. 1
]‫[التاريخ‬
Introduction
Heat conduction is one of three basic types of thermal energy transport
which its conduction, convection and radiation its occur by collision of
microscopic particles and movement of electrons within a body so that when
two things at different temperature touch one another energy dirtily transfer
from the material that have the higher temperature to the material with low
temperature
The objective of the experiment
The aim of this experiment is to measure the thermal conductivity of a solid
cylindrical iron specimen and calculate the thermal conductivity in different
procedures

3. 2
Conductivity
Thermal conductivity can be defined as the rate at which heat is transferred
by conduction through a unit cross-section area of a material, when a
temperature gradient exits perpendicular to the area.
𝑄 = −𝐾𝐴 ∗
∆𝑇
∆𝐿
Where :
Q : Heat transfer (W).
K : Thermal conductivity for specimen, (W/m.°C).
A : Cross-sectional area, (𝑚2
).
∆𝑇
∆𝐿
: Temperature gradient, (°C/m).

4. 3
Device used in the experiment
The specimen is inserted over water reservoir and covered with an insulated
cylinder. Three thermocouples are fixed along the specimen at the points (1,
2 and 3) as shown in the figure. The water temperature recorded with two
thermometers at inlet and outlet also its volume during the experiment time
is recorded. The specimen sides are insulated and the heat transferred along
the specimen to the water.
*solid cylindrical iron specimen has certain dimensions 24 cm in diameter
and 100 mm in length respectively
*The specimen is insulated from other directions so heat flow is one
dimension.
*constant properties (conductivities).

5. 4
Procedure
1- The specimen is heated, using an electrical heater located on the top of the
specimen.
2- The reservoir is supplied with water from a continuous source.
3- The specimen temperature rises and three different locations along the
specimen surface are appointed to measure their temperature values using
thermocouple attached to a selector switch and then to a digital thermometer
until reaching steady state.
4- The heat is transferred from the specimen to the water contained in a
reservoir.
5- The amount of water is measured in couple of minutes (V).
6- Temperature values of the three points and the inlet and outlet water
stream temperature is recorded.
Calculation procedure
The heat transferred to the water from the specimen, the heat
balance can be written as below:
−𝐾𝐴 ∗
∆𝑇
∆𝐿
= 𝑚 𝑐𝑝 ∆𝑇
𝑚 = 𝑣 ∗ 𝜌
Where
K: Thermal conductivity for specimen, (W/m.℃).
A: Cross-sectional area, (𝑚2
).
∆𝑇
∆𝐿
: Temperature gradient, (℃/m)
m: Water mass flow rate, (Kg/s)
cp: Specific heat of water at constant pressure = 4.136 J/kg. ℃.
∆T: is the temperature difference between water inlet and outlet.
V= volume flow rate which is equal to 0.0031 𝑚3
/𝑠

6. 5
 In order to achieved a heat transfer with conduction only the specimen
is insulated completely so that neither convection and nor radiation
heat transfer take place.
Calculations
−𝐾𝐴 ∗
∆𝑇
∆𝐿
= 𝑚 𝑐𝑝 ∆𝑇
According to equation we can compute the conductivity of first specimen by
𝐾1𝐴 ∗
(𝑇1 − 𝑇2)
∆𝐿
= 𝑉 ∗ 𝜌 𝑐𝑝 (𝑇𝑜𝑢𝑡 − 𝑇𝑖𝑛)
𝑘1 ∗
0.0242
4
𝜋 ∗
(66.6 − 63.6925)
0.032
=
0.0031
60
∗ 1000 ∗ 4.136 ∗ (40.3425 − 27.635)
Then k1=66.0646
𝑤
𝑚∗𝑘
conductivity of second specimen
𝐾2𝐴 ∗
(𝑇2 − 𝑇3)
∆𝐿
= 𝑉 ∗ 𝜌 𝑐𝑝 (𝑇𝑜𝑢𝑡 − 𝑇𝑖𝑛)
𝑘2 ∗
0.0242
4
𝜋 ∗
(63.6925 − −60.7825)
0.039
=
0.0031
60
∗ 1000 ∗ 4.136 ∗ (40.3425 − 27.635)
Then k2=80.44709
𝑤
𝑚.𝐾
𝐾3 𝐴 ∗
(𝑇1 − 𝑇3)
∆𝐿
= 𝑉 ∗ 𝜌 𝑐𝑝 (𝑇𝑜𝑢𝑡 − 𝑇𝑖𝑛)
𝑘3
0.0242
4
𝜋 ∗
(66.6 − 60.7825)
0.032
=
0.0031
60
∗ 1000 ∗ 4.136 ∗ (40.3425 − 27.635)
K3=73.2589
𝑤
𝑚.𝑘
.
𝑘𝑎𝑣𝑔 =
𝑘1 + 𝑘2 + 𝑘3
3
=
66.065 + 80.447 + 73.258
3
= 73.256
𝑤
𝑚𝑘
Time (min) Tw in Tw out T1 T2 T3
0 27.6 40.35 66.6 63.7 60.8
5 27.65 40.3 66.65 63.67 60.77
10 27.63 40.37 66.58 63.71 60.76
15 27.66 40.35 66.57 63.69 60.8
Average 27.635 40.3425 66.6 63.6925 60.7825

7. 6
Discussion
1- Using the experimental (T- L) plot and your calculation, find the top
surface temperature and bottom surface temperature of the specimen.
Compare your results.
𝐾𝑎𝑣𝑔𝐴 ∗
(𝑇𝑠 − 𝑇1)
∆𝐿
= 𝑉 ∗ 𝜌 𝑐𝑝 (𝑇𝑜𝑢𝑡 − 𝑇𝑖𝑛)
73.256 ∗
0.0242
4
𝜋 ∗
(𝑇𝑠 − 66.6)
0.015
=
0.0031
60
∗ 1000 ∗ 4.136 ∗ (40.3425 − 27.635)
Ts=67.829 C which is the surface temperature
𝐾𝑎𝑣𝑔𝐴 ∗
(𝑇𝑠 − 𝑇𝑏)
∆𝐿
= 𝑉 ∗ 𝜌 𝑐𝑝 (𝑇𝑜𝑢𝑡 − 𝑇𝑖𝑛)
73.256 ∗
0.0242
4
𝜋 ∗
(67.829 − 𝑇𝑏)
0.1
=
0.0031
60
∗ 1000 ∗ 4.136 ∗ (40.3425 − 27.635)
Tb=59.635 C which is the bottom temperature
2- Can you calculate (K) for two different materials at the same time? Why?
-No, because the two materials have different thermal conductivities and
also different properties.
3- State three applications for conduction in real life.
- in cocking, for example when boiling water.
- In ice melting over surfaces.
- The heat of tee make the cup itself hut.
4- Discuss the main results that you obtained from the experiment.
There is an inverse relationship between the temperature and the distance
and that what we note from our calculations where Ts at surface is more than
Tb which it the bottom temperature.

8. 7